Magnetic vector potential

The magnetostatic magnetic Gauss law is ∇·B=0. On a simply connected regular region (subject to appropriate global topology and boundary conditions), a divergence-free field can be represented locally as the curl of some vector field A. This follows from the vector-calculus solenoidal-field construction; it is not a claim that every global topology admits the same smooth single-valued potential.

Unlike the electrostatic E=−∇V relation, a general B is not a gradient field: the stationary current law allows ∇×B=μ₀J.

Let χ(r) be a sufficiently smooth, single-valued scalar on the considered region. Replace A by A′=A+∇χ. Since the curl of a gradient vanishes, this produces exactly the same B. The value of an individual component of A is therefore gauge dependent.

An open-path integral of A can change under a gauge transformation. For a closed path, ∮∇χ·dℓ=0 when χ is single-valued along the loop, so its A circulation is gauge invariant. Singular gauge functions and multiply connected domains require extra care and cannot be dismissed by a casual global argument.

Use the stationary vacuum source equation ∇×B=μ₀J and substitute B=∇×A. The exact vector identity for a double curl gives ∇×(∇×A)=∇(∇·A)−∇²A. Gauge freedom allows the Coulomb-gauge condition ∇·A=0 for an appropriate localized steady-current solution.

The resulting equation is a componentwise Poisson equation. It is a differential boundary-value problem: specifying only the current without an appropriate decay or boundary prescription does not uniquely specify A, even after a partial gauge choice.

For a current distribution J(r′) localized in a finite region, with sufficient decay at infinity and no imposed harmonic homogeneous solution, use the free-space Green kernel satisfying ∇²[1/(4πR)]=−δ³(r−r′). Apply it to each Cartesian component of the vector Poisson equation.

This gives A as a VECTOR source integral with the same single inverse-distance kernel for every component. For a divergence-free, localized J, integration by parts makes ∇·A=0 when the surface term at infinity vanishes. The formula is not directly an infinity-normalized potential for an ideal infinite line current.

Take ∇ with respect to r, not the primed source coordinate r′. Since J(r′) is independent of r, ∇×[J(r′)/R]=∇(1/R)×J(r′). Because ∇(1/R)=−(r−r′)/R³, antisymmetry of the cross product converts the expression into J×(r−r′)/R³.

The sign is fixed before integration, recovering the same Biot–Savart law as EM-02.02. This is an equivalence within the stationary localized-current model, not a derivation of the full retarded fields of arbitrary time-dependent sources.

For an oriented smooth surface S with closed boundary C, Stokes' theorem gives ∮_C A·dℓ=∫_S(∇×A)·n̂dS=∫_S B·n̂dS. The flux is a physically defined quantity; the regular gauge transformation leaves this closed-loop integral unchanged.

For a uniform B=+B ẑ and a circular disk of radius a with its positive normal +ẑ, its flux is Bπa². This does not assert that A itself is unique or directly measurable in the same manner as B.

Let B=B ẑ be constant. The symmetric-gauge choice is A_sym=(−By/2,Bx/2,0); direct Cartesian differentiation gives (∇×A_sym)_z=∂x(By/2)−∂y(−By/2)=B. A second (Landau) gauge is A_L=(0,Bx,0), which also has curl +B ẑ.

Their difference is ∇χ with χ=Bxy/2. For B=0.200 T, x=0.0300 m, y=0.0400 m, the two A vectors at that single point differ numerically while describing identical magnetic force and flux.

The localized-source Green integral may not converge for an infinite straight filament; instead use its already established field Bφ(s)=μ₀I/(2πs). Choose a purely axial potential A=A_z(s) ẑ. In cylindrical coordinates the azimuthal curl component is (∇×A)_φ=−dA_z/ds.

Integrating with the finite reference A_z(s₀)=0 produces A_z(s)=−μ₀I/(2π)ln(s/s₀) for s>0. An infinity-zero reference is inappropriate because the logarithm diverges as s→∞. A_z's additive constant is gauge-dependent; Bφ is independent of s₀.

Begin by declaring stationary or time-varying sources, vacuum/material model, gauge and boundary conditions. Check whether a candidate A has the desired curl and whether the gauge difference is a true gradient. For a localized steady current, choose a free-space Green solution only when the source integral and its boundary conditions are controlled.

For a circular disk of radius a=0.0500 m in uniform B=0.200 T, Φ_B=Bπa²≈1.57×10⁻³ Wb. The closed-loop integral of either A_sym or A_L equals this flux despite different pointwise potentials. In time-dependent electromagnetic theory, E=−∇V−∂A/∂t and gauge transformations involve V as well; that full relation belongs in EM-03.

1. Compute the curl of A=(−By/2,Bx/2,0) for constant B and confirm it yields B ẑ.
2. Show explicitly that A_L=(0,Bx,0) and A_sym differ by ∇(Bxy/2), and calculate both at B=0.200 T, x=0.0300 m, y=0.0400 m.
3. Use the vector identity and Coulomb gauge to derive ∇²A=−μ₀J from stationary ∇×B=μ₀J.
4. Take the observation-coordinate curl of A(r)=μ₀/(4π)∫J(r′)/|r−r′|d³r′ and recover the signed Biot–Savart kernel.
5. For B=0.200 T through a +z-oriented circular disk a=0.0500 m, compute ∮A·dℓ and its SI units.
6. For a +z infinite filament, integrate Bφ=μ₀I/(2πs) to find an axial A_z(s) vanishing at finite s₀; explain the failure of A_z(∞)=0.

Chapter 05 derives inductance and magnetic energy for linear quasistatic circuits, including self versus mutual inductance, flux linkage, field-energy density, and limits in lossy/dispersive magnetic materials.